Solving Skydiving Problems: Acceleration & Air Resistance

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SUMMARY

The discussion focuses on solving physics problems related to skydiving, specifically the acceleration of sky divers under the influence of air resistance and gravity. The divers, with a total mass of 102.0 kg, experience an upward air resistance force equal to one-fourth of their weight, leading to a net force calculation of 749.7 N. When the parachute is deployed, the divers descend at a constant speed, indicating that the force of air resistance equals their weight, thus resulting in zero acceleration. Participants emphasize understanding net force and its relation to acceleration using Newton's second law.

PREREQUISITES
  • Newton's Second Law of Motion
  • Basic algebra for solving equations
  • Understanding of forces, including weight and air resistance
  • Concept of constant velocity and its implications on acceleration
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  • Study Newton's Second Law in detail, focusing on F=ma
  • Learn about forces acting on objects in free fall and terminal velocity
  • Explore the concept of air resistance and its dependence on speed and surface area
  • Review algebraic manipulation techniques for solving physics equations
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of falling objects and the effects of air resistance on motion.

confusedaboutphysics
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(a) What is the acceleration of two falling sky divers (mass 102.0 kg including parachute) when the upward force of air resistance is equal to one-fourth of their weight?
________ m/s2 (downward)

(b) After popping open the parachute, the divers descend leisurely to the ground at constant speed. What now is the force of air resistance on the sky divers and their parachute? See Fig. 4-34.
_____ N (upward)

i need help on starting this problem...like what equation(s) would i use??

thanks so much!
 
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confusedaboutphysics said:
like what equation(s) would i use??

Instead of immediately looking for an equation, try thinking about the problem.

(a) What is the acceleration of two falling sky divers (mass 102.0 kg including parachute) when the upward force of air resistance is equal to one-fourth of their weight?

If the sky divers are subjected to an upward force equal to one-fourth of their total weight, then what is the net force acting on them?

Once you determine that, then think about what law of physics relates net force (which you will have determined by this time) to acceleration (which is what the question is asking for).

(b) After popping open the parachute, the divers descend leisurely to the ground at constant speed. What now is the force of air resistance on the sky divers and their parachute? See Fig. 4-34.

If the divers' speed is constant, then what is their acceleration? What then can you say about the net force acting on them?
 
can someone tell me if I'm doing this right for part A?

Fnet = 102(9.8) - (.25)(102)(9.8)
Fnet = 999.6 - 249.9
Fnet = 749.7 = ma

so if I'm doing this right..where do i go from here?
 
confusedaboutphysics said:
can someone tell me if I'm doing this right for part A?

Fnet = 102(9.8) - (.25)(102)(9.8)
Fnet = 999.6 - 249.9
Fnet = 749.7 = ma

So far, so good. Just make sure that you understand that you have elected to take the downward direction as positive. Nothing wrong with that, but you'll find that most physics books choose the opposite convention.

so if I'm doing this right..where do i go from here?

What does the question ask you for?
 
so i know I'm trying to find the acceleration..so would i plug in the answer i got for the mass?

Fnet = 749.7 = ma
Fnet = 749.7a?? but how can i solve from here if i don't know the Fnet and I'm looking for a?
 
You do have the net force! You even stated it:

Fnet = 749.7 = ma

So you now know 749.7 = ma or a = 749.7/m

Make sure you understand the definition of net force.

You set 749.7 = m which isn't correct - think of units.
 
Last edited:
thanks tom and cscott! i understand it much better now
 
Not a problem. :smile:
 
confusedaboutphysics said:
Fnet = 749.7 = ma
Fnet = 749.7a??

Have you taken algebra? You can't just move the "a" from one side of the equals sign to the other like that.
 

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